CPTR311 Discrete Structures

1. CPTR311Discrete Structures IntegersReading: Kolman, Section 1.4

2. Divisibility • If one integer, n, divides into a second integer, m, without producing a remainder, then we say that “n divides m”. • Denoted n | m • If one integer, n, does not divide evenly into a second integer, m, i.e., mn produces a remainder, then we say that “n does not divide m” • Denoted n | m

3. Some Properties of Divisibility • If n | m, then there exists a q such that m = qn • The absolute values of both q and n are less than the absolute value of m, i.e., |n| < |m| and |q| < |m| • Examples: 4 | 24: 24 = 46 and both 4 and 6 are less than 24. 5 | 135: 135 = 527 and both 5 and 27 are less than 135 • Simple properties of divisibility • If a | b and a | c, then a | (b + c) • If a | b and a | c, where b > c, then a | (b - c) • If a | b or a | c, then a | bc • If a | b and b | c, then a | c

4. Prime Numbers • A number p is called prime if the only positive integers that divide p are p and 1. • Examples of prime numbers: 2, 3, 5, 7, 11, and 13.

5. Factoring a Number into its Primes • Dividing a number into its multiples over and over again until the multiples cannot be divided any longer shows us that any number can eventually be broken down into prime numbers. • Examples: 9 = 33 = 3224 = 83 = 2223 = 233315 = 3105 = 3335 = 3357 = 3257 • Basically, this means that any number can be broken into multiples of prime numbers.

6. Factoring into Primes (continued) Each row of the table below presents a different number factored into its primes. The numbers in the columns represent the number of each particular prime can be factored out of each original value.

7. Factoring into Primes (continued) • Every positive integer n > 1 can be broken into multiples of prime numbers. • n = p1k1p2k2p3k3p4k4 …psksp1 < p2 < p3 < p4 < …< ps

8. Methods for Factoring • 2 | n  If least significant digit of n is divisible by 2 (i.e., n is even), then 2 divides n • 3 | n  If the sum of all the digits of n down to a single digit equals 3, 6, or 9, then 3 divides n. For example, is 17,587,623 divisible by 3? 1 + 7 + 5 + 8 + 7 + 6 + 2 + 3 = 39 3 + 9 = 12 1 + 2 = 3  YES! 3 divides 17,587,623

9. Methods for Factoring (continued) • Does 7 divide n? • Remove least significant digit (one’s place) from n and multiply it by two. • Subtract the doubled number from the remaining digits. • If result is divisible by 7, then original number was divisible by 7 • Repeat if unable to determine from result.

10. Methods for Factoring (continued) Examples of checking for divisibility by 7 • 1,876  187 – 12 = 175  17 – 10 = 7  • 4,923  492 – 6 = 486  48 – 12 = 36  • 34,461  3,446 – 2 = 3,444 344 – 8 = 336  33 – 12 = 21 

11. Methods for Factoring (continued) • Does 11 divide n? • Starting with the most significant digit of n, adding the first digit, subtracting the next digit, adding the third digit, subtracting the fourth, and so on. If the result is 0 or a multiple of 11, then the original number is divisible by 11. • Repeat if unable to determine from result.

12. Methods for factoring (continued) Examples of checking for divisibility by 11 • 285311670611  2 – 8 + 5 – 3 + 1 – 1 + 6 – 7 + 0 – 6 + 1 – 1 = –11  • 279048  2 – 7 + 9 – 0 + 4 – 8 = 0 

13. Methods for Factoring (continued) • Does 13 divide n? • Delete the last digit (one’s place) from n. • Subtract nine times the deleted digit from the remaining number. • If what is left is divisible by 13, then so is the original number. • Repeat if unable to determine from result.

14. Greatest Common Divisor • If a, b, and k are in Z+, and k | a and k | b, we say that k is a common divisor. • If d is the largest such k, d is called the greatest common divisor (GCD).

15. GCD Example Find the GCD of 540 and 315: • 540 = 22 33  5 • 315 = 32  5  7 • 540 and 315 share the divisors 3, 32, 5, 35, and 325 (Look at it as the number of possible ways to combine 3, 3, and 5) • The largest is the GCD  325 = 45 • 31545 = 7 and 54045=12

16. Euclidean Algorithm • The Euclidean Algorithm is a recursive algorithm that can be used to find GCD (a, b) • It is based on the fact that for any two integers, a > b, there exists a k and r such that: a = kb + r • Since if a | b and a | c, then a | (b + c), then we know that the GCD (a,b) must also divide r. Therefore, the GCD (a,b) = GCD(b,r)

17. Least Common Multiple • If a, b, and k are in Z+, and a | k, b | k, we say that k is a common multiple of a and b. • The smallest such k, call it c, is called the least common multiple or LCM of a and b • We write c = LCM(a,b)

18. Deriving the LCM • We can obtain LCM from a, b • For any integers a and b, we can write a = p1a1 p2a2 …pkak and b = p1b1 p2b2 …pkbk • LCM(a,b) = p1max(a1,b1) p2max(a2,b2) …pkmax(ak,bk)